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Air contains 79% N2 and 21% O2 on a molar basis. Methane (CH4) is burned with 50% excess air than required stoichiometrically. Assuming complete combustion of methane, the molar percentage of N2 in the products is ________________
Correct answer is between '73,74'. Can you explain this answer?
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Air contains 79% N2and 21% O2on a molar basis. Methane (CH4) is burned...
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Air contains 79% N2and 21% O2on a molar basis. Methane (CH4) is burned...
Molar Composition of Air:
- Nitrogen (N2): 79%
- Oxygen (O2): 21%
Molar Composition of Methane Combustion:
- Methane (CH4): 1 mole
Excess Air:
- 50% excess air compared to stoichiometric requirement
Stoichiometric Reaction:
- CH4 + 2O2 -> CO2 + 2H2O
Complete combustion of methane means that all the methane is converted into carbon dioxide (CO2) and water (H2O) with no byproducts.
Molar Stoichiometric Ratio:
- For every 1 mole of methane, 2 moles of oxygen are required for complete combustion.
Excess Air Calculation:
- To find the amount of excess air, we need to calculate the stoichiometric requirement first.
- Since 2 moles of oxygen are required for 1 mole of methane, the stoichiometric requirement would be 2 moles of oxygen per mole of methane.
- With 50% excess air, the total amount of air required would be (100% + 50%) = 150% of the stoichiometric requirement.
- So, the excess air would be 50% of the stoichiometric requirement, which is 0.5 * 2 = 1 mole of oxygen.
Molar Composition of Products:
- With complete combustion of methane, all the methane is converted into carbon dioxide and water.
- The molar composition of the products would be 1 mole of carbon dioxide, 2 moles of water, and the remaining nitrogen from the air.
- Since the molar ratio of nitrogen to oxygen in air is 79:21, the molar ratio of nitrogen to oxygen in the products would also be 79:21.
Molar Percentage of Nitrogen in the Products:
- To find the molar percentage of nitrogen in the products, we need to calculate the total moles of nitrogen in the products and divide it by the total moles of all the products.
- The total moles of nitrogen in the products would be the same as the moles of nitrogen in the air, which is 79% of the total moles of air.
- The total moles of all the products would be the sum of moles of carbon dioxide, water, and nitrogen.
- Since the molar ratio of nitrogen to oxygen in the products is 79:21, the molar percentage of nitrogen in the products would be (79 / (79 + 21)) * 100%.
- Calculating this, we get (79 / 100) * 100% = 79%.
Therefore, the molar percentage of nitrogen in the products is approximately 79%.
- Nitrogen (N2): 79%
- Oxygen (O2): 21%
Molar Composition of Methane Combustion:
- Methane (CH4): 1 mole
Excess Air:
- 50% excess air compared to stoichiometric requirement
Stoichiometric Reaction:
- CH4 + 2O2 -> CO2 + 2H2O
Complete combustion of methane means that all the methane is converted into carbon dioxide (CO2) and water (H2O) with no byproducts.
Molar Stoichiometric Ratio:
- For every 1 mole of methane, 2 moles of oxygen are required for complete combustion.
Excess Air Calculation:
- To find the amount of excess air, we need to calculate the stoichiometric requirement first.
- Since 2 moles of oxygen are required for 1 mole of methane, the stoichiometric requirement would be 2 moles of oxygen per mole of methane.
- With 50% excess air, the total amount of air required would be (100% + 50%) = 150% of the stoichiometric requirement.
- So, the excess air would be 50% of the stoichiometric requirement, which is 0.5 * 2 = 1 mole of oxygen.
Molar Composition of Products:
- With complete combustion of methane, all the methane is converted into carbon dioxide and water.
- The molar composition of the products would be 1 mole of carbon dioxide, 2 moles of water, and the remaining nitrogen from the air.
- Since the molar ratio of nitrogen to oxygen in air is 79:21, the molar ratio of nitrogen to oxygen in the products would also be 79:21.
Molar Percentage of Nitrogen in the Products:
- To find the molar percentage of nitrogen in the products, we need to calculate the total moles of nitrogen in the products and divide it by the total moles of all the products.
- The total moles of nitrogen in the products would be the same as the moles of nitrogen in the air, which is 79% of the total moles of air.
- The total moles of all the products would be the sum of moles of carbon dioxide, water, and nitrogen.
- Since the molar ratio of nitrogen to oxygen in the products is 79:21, the molar percentage of nitrogen in the products would be (79 / (79 + 21)) * 100%.
- Calculating this, we get (79 / 100) * 100% = 79%.
Therefore, the molar percentage of nitrogen in the products is approximately 79%.
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Air contains 79% N2and 21% O2on a molar basis. Methane (CH4) is burned with 50% excess air than required stoichiometrically. Assuming complete combustion of methane, the molar percentage of N2in the products is ________________Correct answer is between '73,74'. Can you explain this answer?
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Air contains 79% N2and 21% O2on a molar basis. Methane (CH4) is burned with 50% excess air than required stoichiometrically. Assuming complete combustion of methane, the molar percentage of N2in the products is ________________Correct answer is between '73,74'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Air contains 79% N2and 21% O2on a molar basis. Methane (CH4) is burned with 50% excess air than required stoichiometrically. Assuming complete combustion of methane, the molar percentage of N2in the products is ________________Correct answer is between '73,74'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Air contains 79% N2and 21% O2on a molar basis. Methane (CH4) is burned with 50% excess air than required stoichiometrically. Assuming complete combustion of methane, the molar percentage of N2in the products is ________________Correct answer is between '73,74'. Can you explain this answer?.
Air contains 79% N2and 21% O2on a molar basis. Methane (CH4) is burned with 50% excess air than required stoichiometrically. Assuming complete combustion of methane, the molar percentage of N2in the products is ________________Correct answer is between '73,74'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Air contains 79% N2and 21% O2on a molar basis. Methane (CH4) is burned with 50% excess air than required stoichiometrically. Assuming complete combustion of methane, the molar percentage of N2in the products is ________________Correct answer is between '73,74'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Air contains 79% N2and 21% O2on a molar basis. Methane (CH4) is burned with 50% excess air than required stoichiometrically. Assuming complete combustion of methane, the molar percentage of N2in the products is ________________Correct answer is between '73,74'. Can you explain this answer?.
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Air contains 79% N2and 21% O2on a molar basis. Methane (CH4) is burned with 50% excess air than required stoichiometrically. Assuming complete combustion of methane, the molar percentage of N2in the products is ________________Correct answer is between '73,74'. Can you explain this answer?, a detailed solution for Air contains 79% N2and 21% O2on a molar basis. Methane (CH4) is burned with 50% excess air than required stoichiometrically. Assuming complete combustion of methane, the molar percentage of N2in the products is ________________Correct answer is between '73,74'. Can you explain this answer? has been provided alongside types of Air contains 79% N2and 21% O2on a molar basis. Methane (CH4) is burned with 50% excess air than required stoichiometrically. Assuming complete combustion of methane, the molar percentage of N2in the products is ________________Correct answer is between '73,74'. Can you explain this answer? theory, EduRev gives you an
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